If `f(x)= int_(x)^(x^(2)) (dt)/(1+ t^(3))`, then f'(2)=

Options

A. `-(56)/(585)`
B. `-(29)/(585)`
C. `(101)/(585)`
D. none

Answer

To find \( f'(2) \) for the function \( f(x) = \int_{x}^{x^2} \frac{dt}{1 + t^3} \), we will use the Leibniz rule for differentiating under the integral sign.

### Step 1: Differentiate \( f(x) \)

Using the Leibniz rule, we differentiate \( f(x) \):

\[ f'(x) = \frac{d}{dx} \left( \int_{x}^{x^2} \frac{dt}{1 + t^3} \right) = \frac{1}{1 + (x^2)^3} \cdot \frac{d}{dx}(x^2) - \frac{1}{1 + x^3} \cdot \frac{d}{dx}(x) \]

### Step 2: Calculate the derivatives of the limits

The derivative of the upper limit \( x^2 \) is:

\[ \frac{d}{dx}(x^2) = 2x \]

The derivative of the lower limit \( x \) is:

\[ \frac{d}{dx}(x) = 1 \]

### Step 3: Substitute the derivatives into the expression

Substituting these derivatives back into the expression for \( f'(x) \):

\[ f'(x) = \frac{1}{1 + x^6} \cdot 2x - \frac{1}{1 + x^3} \cdot 1 \]

### Step 4: Simplify the expression

Now we can simplify:

\[ f'(x) = \frac{2x}{1 + x^6} - \frac{1}{1 + x^3} \]

### Step 5: Evaluate \( f'(2) \)

Now we substitute \( x = 2 \):

\[ f'(2) = \frac{2 \cdot 2}{1 + 2^6} - \frac{1}{1 + 2^3} \]

Calculating \( 2^6 \) and \( 2^3 \):

\[ 2^6 = 64 \quad \text{and} \quad 2^3 = 8 \]

Substituting these values:

\[ f'(2) = \frac{4}{1 + 64} - \frac{1}{1 + 8} \]

This simplifies to:

\[ f'(2) = \frac{4}{65} - \frac{1}{9} \]

### Step 6: Find a common denominator and combine

The common denominator for \( 65 \) and \( 9 \) is \( 585 \):

\[ f'(2) = \frac{4 \cdot 9}{585} - \frac{1 \cdot 65}{585} = \frac{36 - 65}{585} = \frac{-29}{585} \]

### Final Answer

Thus, we have:

\[ f'(2) = -\frac{29}{585} \]